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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 501. |
Isotonic solutions must have the sameA. solteB. densityC. elevation in boilling pointD. depression in freezing point |
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Answer» Correct Answer - `(c,d)` istonic solutions have same molar concentration and therefore, same colligative properties. Options (c,d) are both correct. |
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| 502. |
What are the Types of azeotropes? Explain with example. |
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Answer» Types of azeotropes: Minimum boiling azeotrope The non- ideal solutions showing positive deviation form minimum boiling azeotrope at a specific composition. Example: 95% ethanol and 5% water (by volume) Ethanol = 351.3 K , Water = 373 K, Azeotrope = 351.1 K Maximum boiling azeotrope The non- ideal solutions showing negative deviation form maximum boiling azeotrope at a specific composition. Example : 68% Nitric acid and 32% water (by mass) Nitric acid = 359 k, water = 373 k, Azeotrope = 393.5 K |
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| 503. |
What are isotonic solutions? |
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Answer» Two solutions having same osmotic pressure at a given temperature are called isotonic solutions. When such solutions are separated by semipermeable membrane no osmosis occurs between them. |
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| 504. |
Define mole fraction. |
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Answer» Mole fraction: It is defined as the ratio of number of moles of the component to that of the total number of moles of all the components .Commonly used symbol for mole fraction is x and subscript used on the right hand side of x denotes the component. For example, in a binary mixture, if the number of moles of A and B are nA and nB respectively, the mole fraction of A will be xA = nA / nA+nB. |
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| 505. |
Non ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example of each type. |
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Answer» The cause for these deviations lies in the nature of interactions at the molecular level. In case of positive deviation from Raoult‘s law, A-B interactions are weaker than those between A-A or B-B, i.e., in this case the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules. This means that in such solutions, molecules of A (or B) will find it easier to escape than in pure state. This will increase the vapour pressure and result in positive deviation. Mixtures of ethanol and acetone behave in this manner. In case of negative deviations from Raoult‘s law, the intermolecularattractive forces between A-A and B-B are weaker than those between A-B and lead to decrease in vapour pressure resulting in negative deviations. An example of this type is a mixture of phenol and aniline. |
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| 506. |
What is Azeotropes ? |
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Answer» Azeotropes are binary solutions (liquid mixtures) having the same composition in liquid and vapour phase and it is not possible to separate the components of an azeotrope by fractional distillation. |
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| 507. |
Among the following substances, the lowest vapour pressure is exerted byA. WaterB. MercuryC. KeroseneD. Rectified spirit |
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Answer» Correct Answer - B Nonvolatile substance has V.P. |
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| 508. |
`18`g of glucose `(C_(6)H_(12)O_(6))`is added to `178.2` g of water. The vapour pressure of water for this aqueous solution at `100^@C`is :A. `752.40`torrB. `759.00`torrC. `7.60`torrD. `76.00`torr |
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Answer» Correct Answer - A `(P^(@)+P_(s))/(P_(S))=(wxxM)/(mxxW)` at.b.pt.`P^(@)=760mm` `(760-P_(s))/(P_(s))=(18xx18)/(180xx178.2)` `P_(s)=752.40` torr |
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| 509. |
The vapour pressure of pure liquid solvent A is `0.80`atm.When a non volatile substances B is added to the solvent, its vapour pressure drops to `0.60`atm. Mole fraction of the components B in the solution is:A. `0.50`B. `0.25`C. `0.75`D. gives data is not sufficient |
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Answer» Correct Answer - B `(P^(@) -P_(s))/(P^(@)) = X_("solute")` `X_("solute") = 0.25` |
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| 510. |
The vapour pressure of pure liquid solvent A is `0.80`atm.When a non volatile substances B is added to the solvent, its vapour pressure drops to `0.60`atm. Mole fraction of the components B in the solution is:A. `0.50`B. `0.25`C. `0.75`D. `0.40` |
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Answer» Correct Answer - B Acc. To R.L.V.P. `(cancelOP)/(P_(0))=X_(cancelB)` `X_(B)(0.80-0.60)/(0.80)=(0.20)/(0.80)=0.25.` |
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| 511. |
What weight of 60% pure NaOH is required to neutralise 100 ml `(M)/(1o)H_(2)SO_(4)` solution ? |
| Answer» Correct Answer - 1.33 g | |
| 512. |
Which one of the following statement is correct.A. Brownian movement is more pronounced for smaller particles than for bigger onesB. Sols of metal sulphides are lyophilicC. Schulze-Hardy law states, the bigger the size of the ion, the greater is its coagulating powerD. One would expect charcoal to adsorb hydrogen gas more strongly than chlorine |
| Answer» Correct Answer - A | |
| 513. |
If `Cl_(2)` gas is enclosed in presence of powdered charcoal in a closed vessel, the pressure of the gas decreases. It is because :A. the gas moleciles are adsorbed at the surfaceB. the gas molecules concentrate at the surface of the charcoalC. the gas molecules are adsorbed at the surfaceD. the gas molecules are desorbed by the surface |
| Answer» Correct Answer - B::C | |
| 514. |
The potential differnce between the fixed particles layer and the diffused layer having opposite charge id called :A. Water potentialB. Zeta potentialC. Electrode potentialD. None of these |
| Answer» Correct Answer - B | |
| 515. |
Hydrolysis of ester in catalysed by acid. Rates of hydrolysis of ester were obtained initially and after `50%` ester has been hydrolused as `R_(0)` and `R_(50)` then (same temp.)A. `R_(0)=R_(50)`B. `R_(0) lt R_(50)`C. `R_(0) gt R_(50)`D. Cannot be determined |
| Answer» Correct Answer - B | |
| 516. |
Finely divided catalyst has greater surface area and has greater catalytic activity than the compact solid. If a total surface area of 6291456 `cm^(2)` is required for adsorption in a catalytic gaseous reaction, then how many splits should be made in a cube of exactly 1 cm in length to achieve required surface area? [Given : One split of a cube gives eight cubes of same size]A. `60`B. `80`C. `20`D. `22` |
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Answer» Correct Answer - C Total surface area of eight cubes `=8xx6xx(1/2xx1/2)` Apply the formula Surface area on a split of a cube `=8^(n)xx6xx(1/2)^(2n)` `6291456=8^(n)xx6xx(1/2)^(2n)` |
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| 517. |
Volume of `N_(2)` at 1 atm, 273 K required to form a monolayer on the surface of iron catalyst is `8.15ml//gm` of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies `16 xx 10^(-22)m^(2)`? [Take : `N_(A)=6xx10^(23)`]A. `16xx10^(-16) cm^(2)`B. `0.35 m^(2)//g`C. `39 m^(2)//g`D. `22400 cm^(2)` |
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Answer» Correct Answer - B The volume of `N_(2)` at STP required to cover the ion surface with monolayer `=8.15 ml gm^(-1)` Area occupied by single molecule `=16xx10^(-18) cm^(2)` `22400` ml of `N_(2)` at STP contains `=N_(A)` molecule of `N_(2)` `:. 8.15 ……………. =(8.15xxN_(A))/(22400)=2.19xx10^(20)` molecule of `N_(2)` Area occupied by `2.19xx10^(20)` molecule of `N_(2)=2.19xx10^(20)xx16xx10^(-18) cm^(2)=35.06xx10^(2) cm^(2)` surface area of the iron adsorbed `=0.35 m^(2) gm^(-1)` In short `A=("Volume covered by the" N_(2) "molecule"xxN_(A)xx"Area occupied by single molecule")/22400` |
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| 518. |
For a solution if `p_(A)^(@)=600" mm Hg",p_(B)^(@)=840" mm Hg"` under atmospheric conditions and vapour pressure of solution is 1 atm then find (i) Composition of solution (ii) Composition of vapour in equilibrium wiht solution |
| Answer» `760=600 x_(A)+840(1-x_(A))` | |
| 519. |
Vapour pressure of water is 360 mm Hg, how much urea should be added to 200 mL water to reduce its vapour pressure by 0.5% (Molecular wt. of urea =60)`(100)/(3)g)` |
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Answer» Correct Answer - 3.33 g `(0.005p_(A)^(@))/(p_(A)^(@))=((x)/(60))/((x)/(60)+(200)/(18))` |
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| 520. |
The degree of association `(alpha)` is given by the expressionA. `alpha=(i-1)/(1/n-1)`B. `alpha=(i-1)/(n-1)`C. `alpha=(1-i)/(1/n-1)`D. `alpha=(1-i)/(n-1)` |
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Answer» Correct Answer - 3 Degree of association `(alpha)` is the fraction of one mole of solute which associates to form bigger particle:- `nA hArr (A)_(n)` moles 1 mol 0 mol before assocation Moles `(1 - alpha)` mol `((alpha)/(n))` mole after associates `i=("Molar after association")/("Moles before association")` `=((1-alpha)+(alpha/n))/1=(1+(1/n-1)alpha)/1` or `(1/n-1)alpha=i-1` `alpha=(i-1)/((1/n-1))` For dissociation `alpha=(i-1)/(n-1)` |
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| 521. |
Among the following, the solution which shows the highest osmotic pressure isA. `0.04 M FeSO_(4).(NH_(4))_(2)SO_(4).H_(2)O`B. `0.05 M K_(4)[Fe(CN)_(6)]`C. `0.05 M k_[Cu(CN)_(4)]`D. `0.05 M Al(NO_(3))_(3)` |
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Answer» Correct Answer - 2 `pi=I CRT` At a given temperature, solution having the highest value of iC has the gighest osmotic pressure. `K_(4)[Fe(CN)_(6)] rarr 4K^(+)+[Fe(CN)_(6)]^(4-)` `iC=(5)(0.05)=0.25` `FeSO_(4). (NH_(4))_(2)SO_(4).6H_(2)Ooverset(H_(2)O)(rarr) Fe^(2+)+2NH_(4)^(+)+2SO_(4)^(2-)` `iC=(5)(0.04)` `=0.2` `K_(3)[Cu(CN)_(4)]overset(H_(2)O)(rarr) 3K^(+)+[Cu(CN)_(4)]^(3-)` `iC=(4)(0.05)` `=0.2` `Al(NO_(3))_(3)overset(H_(2)O)(rarr)Al^(3+)+3NO_(3)^(-)` `iC=(4)(0.05)` `=0.2` |
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| 522. |
Diluting a solution: How would you prepare 500.0 mL of 0.2500 M NaOH solution starting from a concentration of 1.00 M? Strategy: The problem gives initial and final concentration `(M_(i) and M_(f))` and final volume `(V_(f))` and asks for the initial volume `(V_(i))` that we need to dilute. Write the dilute formula (2.17) and rearrange it to give the initial volume. |
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Answer» According to Equation `(2.17)` `M_(i)V_(i)=M_(f)V_(f)` or `V_(i)=(M_(f)V_(f))/V_(i)` Substituting the known values into the right side of the equation: `V_(i)=((0.2500 M)(500.0 mL))/((1.000 M))` `=125.0 ML` In practice, bilution is carried out as follows: the valume to be diluted is withdrawn using a calibrated tube called a pipet, placed in an empty volumetric flask of the chosen volume, and diluted to the calibration mark on the flask. In this particle case, we need to place `125.0 mL` of `1.000 M NaOH` solution in a `500.0 mL` volumeritc flask and fill to the mark with water. |
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| 523. |
What are Reverse osmosis? Give the uses. |
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Answer» When a pressure larger than osmotic pressure is applied to the solution side, the pure solvent flows out of solution to solvent side through semi permeable membrane this process is called reverse osmosis. Use: Reverse osmosis is used to desalination of sea water (into drinking water) |
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| 524. |
A 25% alcohol solution means ........ (a) 25 ml of alcohol in. 100 ml of water (b) 25 ml of alcohol in 25 ml of water (c) 25 ml of alcohol in 75 ml of water (d) 75 ml of alcohol in 25 ml of water. |
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Answer» (c) 25 ml of alcohol in 75 ml of water |
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| 525. |
Match the following |
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Answer» Correct Answer - A(s), B(r ), C(q), D(p) |
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| 526. |
Deliquescence is due to: (a) Strong affinity to water (b) Less affinity to water (c) Strong hatred to water (d) Inertness to water |
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Answer» (a) Strong affinity to water |
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| 527. |
Match the following |
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Answer» Correct Answer - A(q), B(p ), C(s), D(r ) |
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| 528. |
Match the following |
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Answer» Correct Answer - A(q,s), B(q,s ), C(p,r), D(p,r ) |
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| 529. |
Which characterises the weak intermolecular forces of attraction in a liquid?A. High boiling pointB. High vapour pressureC. High critical temperatureD. High heat of vaporization |
| Answer» Due to weak force of attraction more vapour will be formed so vapour pressure will be high. | |
| 530. |
Match the following |
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Answer» Correct Answer - A(q), B(p,r,t ), C(q,s), D(p ) |
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| 531. |
Let , solute - solvent intermolecular attraction force `=F_(A)` solute - solute intermolecular attraction force =` F_(B)` solvent - solvent intermolecular attraction force `=F_(C)` Match the following |
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Answer» Correct Answer - A(q,r), B(p,t ), C(q,s), D(r ) |
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| 532. |
Define : (1) Solvent (2) Solute. |
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Answer» (1) Solvent : The component in which solution formation takes place and which constitutes larger proportion of a solution is called solvent. For example, in an aqueous solution of sugar, water is the solvent. (2) Solute : In a solution the component which dissolves and constitutes smaller proportion of a solution is called a solute. For example, in a sugar solution, sugar is the solute. |
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| 533. |
Identify the correct statement regarding enzymes.A. Enzymes are specific biological catalysts that can normally function at very high tempt. `(T~~1000 K)`B. Enzymes are normally heterogeneous catalysis that very specific in actionC. Enzymes are specific biological catalysts that can not be poisonedD. Enzymes are non-biological catalysts. |
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Answer» Correct Answer - B Enzymes are highly specific heterogeneous catalyst. |
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| 534. |
The size of colloidal particles is :A. `1-10 Å`B. `20-50 Å`C. `10-100 Å`D. `1-200 Å` |
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Answer» Correct Answer - B::C Size of colloidal particle is `10Å` to `10^(4) Å` |
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| 535. |
Which of the following anions will have minimum flocculation value for the ferric oxide solution?A. `Cl^(-)`B. `Br^(-)`C. `SO_(4)^(2-)`D. `[Fe(CN)_(6)]^(3-)` |
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Answer» Correct Answer - D Higher the charge on coagulating ion, lesser tha flocculation value. |
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| 536. |
Which is an example of coagulation?A. Curdling of milkB. purification of water by addition of alumC. formation of deltas at the river bedsD. All the three are example of coagulation |
| Answer» Correct Answer - D | |
| 537. |
Which of the following statement is correct?A. The efficiency of a heterogeneous catalyst depends upon its surface area.B. Catalyst operates by providing alternate path for the reaction that involves a lower activation energyC. Catalyst lowers the energy of activation of the forward direction without affecting the energy of activation of the backward direction.D. Catalyst does not affect the enthalpy change of the reaction. |
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Answer» Correct Answer - A::B::C Efficiency of a heterogeneous catalyst increases with its surface area. Catalyst provides a path of lower activation energy but enthalpy of reaction is not affected. |
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| 538. |
Which of the following statement is (are) true?A. The concentration of a homogeneous catalyst may appear in the rate expression.B. A catalyst is always consumed in the reactionC. A catalyst must always must always be in the same phase as the reactantsD. None of these |
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Answer» Correct Answer - A Catalyst may appear in rate expression. But it is not consumed in the reaction. |
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| 539. |
The colloidal particles are electrically charged as a indicated by their migration towards cathode or anode under the applied electric field. In a particular colloidal system, all particles carry either positive charge or negative charge. The electric charge on colloidal particles orginate in several ways. According to preferential adsorption theory, the freshly obtained precipitate particles adsorb ions from the dispersion medium, which are common to their lattice and acquire the charge of adsorbed ions. For example, For example, freshly obtained `Fe(OH)_(3)` precipitated is dispersed, by a little `FeCl_(3)`, into colloidal solution owing to the adsorption of `Fe^(3+)` ions in preference. Thus sol particles will be positively charged. In some cases the colloidal particles are aggregates of cations or anions having ampiphilic character. When the ions posses hydrophobic part (hydrocarbon end) as well as hydrophilic part (polar end group), they undergo association in aqueous solution to form particles having colloidal size. The formation of such particles, called micelles plays a very important role in the solubilization of water insoluble substances, (hydrocarbon, oils, fats, grease etc.). In micelles, the polar end groups are directed towards water and the hydrocarbon ends into the centre. The charge on sol particles of proteins depends on the pH. At low pH, the basic group of protein molecule is ionized (protonated) and at higher pH (alkaline medium), the acidic group is ionized. At isoelectric pH, characteristic to the protein, both basix and acidic groups are equally ionized. The stability of colloidal solution is attributed largely to the electric charge of the dispersed particles. This charge causes them to be coagulated or precipitated. On addition of small amount of electrolytes, the ions carrying oppiste charge are adsorbed by sol particles resulting in the neutralization of their charge. When the sol particles either with no charge or reduced charge, come closer due to Brownian movement, they coalesce to form bigger particles resulting in their separation from the dispersion medium. This is what is called coagulating or precipitation of the colloidal solution. The coagulating power of the effective ion, which depend on its charge, is expressed in terms of its coagulating value, defined as its minimum concentration (m mol/L) needed to precipitate a given sol. How would you obtain a sol of AgI, the particles of which migrate towards cathode under the electric field?A. By adding you obtain of KI to `AgNO_(3)` solutionB. By adding little excess of `AgNO_(3)` to KI solutionC. By mixing equal volumes of `0.010 M AgNO_(3)` and `0.010 M` KID. None of these |
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Answer» Correct Answer - B We want to prepare sol of AgI having positively charged particles, so a little excess of `Ag^(+)` should be added to KI. |
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| 540. |
The colloidal particles are electrically charged as a indicated by their migration towards cathode or anode under the applied electric field. In a particular colloidal system, all particles carry either positive charge or negative charge. The electric charge on colloidal particles orginate in several ways. According to preferential adsorption theory, the freshly obtained precipitate particles adsorb ions from the dispersion medium, which are common to their lattice and acquire the charge of adsorbed ions. For example, For example, freshly obtained `Fe(OH)_(3)` precipitated is dispersed, by a little `FeCl_(3)`, into colloidal solution owing to the adsorption of `Fe^(3+)` ions in preference. Thus sol particles will be positively charged. In some cases the colloidal particles are aggregates of cations or anions having ampiphilic character. When the ions posses hydrophobic part (hydrocarbon end) as well as hydrophilic part (polar end group), they undergo association in aqueous solution to form particles having colloidal size. The formation of such particles, called micelles plays a very important role in the solubilization of water insoluble substances, (hydrocarbon, oils, fats, grease etc.). In micelles, the polar end groups are directed towards water and the hydrocarbon ends into the centre. The charge on sol particles of proteins depends on the pH. At low pH, the basic group of protein molecule is ionized (protonated) and at higher pH (alkaline medium), the acidic group is ionized. At isoelectric pH, characteristic to the protein, both basix and acidic groups are equally ionized. The stability of colloidal solution is attributed largely to the electric charge of the dispersed particles. This charge causes them to be coagulated or precipitated. On addition of small amount of electrolytes, the ions carrying oppiste charge are adsorbed by sol particles resulting in the neutralization of their charge. When the sol particles either with no charge or reduced charge, come closer due to Brownian movement, they coalesce to form bigger particles resulting in their separation from the dispersion medium. This is what is called coagulating or precipitation of the colloidal solution. The coagulating power of the effective ion, which depend on its charge, is expressed in terms of its coagulating value, defined as its minimum concentration (m mol/L) needed to precipitate a given sol. Which of the following ions would have the minimum coagulating value for sol obtained on peptizing `Sn(OH)_(4)` by little NaOH solution?A. `Cl^(-)`B. `SO_(4)^(2-)`C. `K^(+)`D. `Ba^(2+)` |
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Answer» Correct Answer - D Minimum coagulating value will be for the ion with maximum charge and since the sol particles are negatively charged, hence positively charged ions are required for coagulations |
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| 541. |
The colloidal particles are electrically charged as a indicated by their migration towards cathode or anode under the applied electric field. In a particular colloidal system, all particles carry either positive charge or negative charge. The electric charge on colloidal particles orginate in several ways. According to preferential adsorption theory, the freshly obtained precipitate particles adsorb ions from the dispersion medium, which are common to their lattice and acquire the charge of adsorbed ions. For example, For example, freshly obtained `Fe(OH)_(3)` precipitated is dispersed, by a little `FeCl_(3)`, into colloidal solution owing to the adsorption of `Fe^(3+)` ions in preference. Thus sol particles will be positively charged. In some cases the colloidal particles are aggregates of cations or anions having ampiphilic character. When the ions posses hydrophobic part (hydrocarbon end) as well as hydrophilic part (polar end group), they undergo association in aqueous solution to form particles having colloidal size. The formation of such particles, called micelles plays a very important role in the solubilization of water insoluble substances, (hydrocarbon, oils, fats, grease etc.). In micelles, the polar end groups are directed towards water and the hydrocarbon ends into the centre. The charge on sol particles of proteins depends on the pH. At low pH, the basic group of protein molecule is ionized (protonated) and at higher pH (alkaline medium), the acidic group is ionized. At isoelectric pH, characteristic to the protein, both basix and acidic groups are equally ionized. The stability of colloidal solution is attributed largely to the electric charge of the dispersed particles. This charge causes them to be coagulated or precipitated. On addition of small amount of electrolytes, the ions carrying oppiste charge are adsorbed by sol particles resulting in the neutralization of their charge. When the sol particles either with no charge or reduced charge, come closer due to Brownian movement, they coalesce to form bigger particles resulting in their separation from the dispersion medium. This is what is called coagulating or precipitation of the colloidal solution. The coagulating power of the effective ion, which depend on its charge, is expressed in terms of its coagulating value, defined as its minimum concentration (m mol/L) needed to precipitate a given sol. A gelatin sol at pH less than the isoelectric value is subjected to an electric field. The sol particles migrate toward :A. AnodeB. CathodeC. Both anode and cathodeD. Neither anode no cathode |
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Answer» Correct Answer - B At low pH the basic group will be ionized (protonated) so will have positive charge and hence sol particles wll move towards cathode. |
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| 542. |
The colloidal particles are electrically charged as a indicated by their migration towards cathode or anode under the applied electric field. In a particular colloidal system, all particles carry either positive charge or negative charge. The electric charge on colloidal particles orginate in several ways. According to preferential adsorption theory, the freshly obtained precipitate particles adsorb ions from the dispersion medium, which are common to their lattice and acquire the charge of adsorbed ions. For example, For example, freshly obtained `Fe(OH)_(3)` precipitated is dispersed, by a little `FeCl_(3)`, into colloidal solution owing to the adsorption of `Fe^(3+)` ions in preference. Thus sol particles will be positively charged. In some cases the colloidal particles are aggregates of cations or anions having ampiphilic character. When the ions posses hydrophobic part (hydrocarbon end) as well as hydrophilic part (polar end group), they undergo association in aqueous solution to form particles having colloidal size. The formation of such particles, called micelles plays a very important role in the solubilization of water insoluble substances, (hydrocarbon, oils, fats, grease etc.). In micelles, the polar end groups are directed towards water and the hydrocarbon ends into the centre. The charge on sol particles of proteins depends on the pH. At low pH, the basic group of protein molecule is ionized (protonated) and at higher pH (alkaline medium), the acidic group is ionized. At isoelectric pH, characteristic to the protein, both basix and acidic groups are equally ionized. The stability of colloidal solution is attributed largely to the electric charge of the dispersed particles. This charge causes them to be coagulated or precipitated. On addition of small amount of electrolytes, the ions carrying oppiste charge are adsorbed by sol particles resulting in the neutralization of their charge. When the sol particles either with no charge or reduced charge, come closer due to Brownian movement, they coalesce to form bigger particles resulting in their separation from the dispersion medium. This is what is called coagulating or precipitation of the colloidal solution. The coagulating power of the effective ion, which depend on its charge, is expressed in terms of its coagulating value, defined as its minimum concentration (m mol/L) needed to precipitate a given sol. When 9.0 ml of arsenious sulphide sol and 1.0 ml of `0.09 M BaCl_(2)` are mixed, turbidity due to precipitation just appears after 2 hours. The effective ion and its coagulating value are respectively :A. `Cl^(-), 10 m mol//L`B. `Cl^(-), 20 m mol//L`C. `Ba^(2+), 10 m mol//L`D. `Ba^(2+), 20 m mol//L` |
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Answer» Correct Answer - C Conc. Of `Ba^(2+)=10^(-4)/(10xx10^(-3)) M=10^(-2) M =10 mmole//L` |
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| 543. |
Select correct statement (s):A. hydrophilic colloid is a colloid in which there is a strong attraction between the dispersed phase and waterB. hydrophobic colloid is a colloid in which there is a lack of attraction between the dispersed phase and waterC. hydrophobic sols are often formed when a solid crystallises rapidly from a chemical reaction or a supersaturated solutionD. all of the above |
| Answer» Correct Answer - D | |
| 544. |
The colloidal particles are electrically charged as a indicated by their migration towards cathode or anode under the applied electric field. In a particular colloidal system, all particles carry either positive charge or negative charge. The electric charge on colloidal particles orginate in several ways. According to preferential adsorption theory, the freshly obtained precipitate particles adsorb ions from the dispersion medium, which are common to their lattice and acquire the charge of adsorbed ions. For example, For example, freshly obtained `Fe(OH)_(3)` precipitated is dispersed, by a little `FeCl_(3)`, into colloidal solution owing to the adsorption of `Fe^(3+)` ions in preference. Thus sol particles will be positively charged. In some cases the colloidal particles are aggregates of cations or anions having ampiphilic character. When the ions posses hydrophobic part (hydrocarbon end) as well as hydrophilic part (polar end group), they undergo association in aqueous solution to form particles having colloidal size. The formation of such particles, called micelles plays a very important role in the solubilization of water insoluble substances, (hydrocarbon, oils, fats, grease etc.). In micelles, the polar end groups are directed towards water and the hydrocarbon ends into the centre. The charge on sol particles of proteins depends on the pH. At low pH, the basic group of protein molecule is ionized (protonated) and at higher pH (alkaline medium), the acidic group is ionized. At isoelectric pH, characteristic to the protein, both basix and acidic groups are equally ionized. The stability of colloidal solution is attributed largely to the electric charge of the dispersed particles. This charge causes them to be coagulated or precipitated. On addition of small amount of electrolytes, the ions carrying oppiste charge are adsorbed by sol particles resulting in the neutralization of their charge. When the sol particles either with no charge or reduced charge, come closer due to Brownian movement, they coalesce to form bigger particles resulting in their separation from the dispersion medium. This is what is called coagulating or precipitation of the colloidal solution. The coagulating power of the effective ion, which depend on its charge, is expressed in terms of its coagulating value, defined as its minimum concentration (m mol/L) needed to precipitate a given sol. Under the influence of an electric field, the particles in a sol migrate towards cathode. The coagulation of the same sol is studied using NaCl, `Na_(2)SO_(4)` and `Na_(3)PO_(4)` solutions. Their coagulating values will be in the order :A. `NaCl gt Na_(2)SO_(4) gt Na_(3)PO_(4)`B. `Na_(2)SO_(4) gt Na_(3)PO_(4)` gt NaCl`C. `Na_(3)PO_(4) gt Na_(2)SO_(4) gt NaCl`D. `Na_(2)SO_(4) gt NaCl gt Na_(3)PO_(4)` |
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Answer» Correct Answer - A We want to prepare sol of AgI having positively charged particles, so a little excess of `Ag^(+)` should be added to KI. |
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| 545. |
The colloidal particles are electrically charged as a indicated by their migration towards cathode or anode under the applied electric field. In a particular colloidal system, all particles carry either positive charge or negative charge. The electric charge on colloidal particles orginate in several ways. According to preferential adsorption theory, the freshly obtained precipitate particles adsorb ions from the dispersion medium, which are common to their lattice and acquire the charge of adsorbed ions. For example, For example, freshly obtained `Fe(OH)_(3)` precipitated is dispersed, by a little `FeCl_(3)`, into colloidal solution owing to the adsorption of `Fe^(3+)` ions in preference. Thus sol particles will be positively charged. In some cases the colloidal particles are aggregates of cations or anions having ampiphilic character. When the ions posses hydrophobic part (hydrocarbon end) as well as hydrophilic part (polar end group), they undergo association in aqueous solution to form particles having colloidal size. The formation of such particles, called micelles plays a very important role in the solubilization of water insoluble substances, (hydrocarbon, oils, fats, grease etc.). In micelles, the polar end groups are directed towards water and the hydrocarbon ends into the centre. The charge on sol particles of proteins depends on the pH. At low pH, the basic group of protein molecule is ionized (protonated) and at higher pH (alkaline medium), the acidic group is ionized. At isoelectric pH, characteristic to the protein, both basix and acidic groups are equally ionized. The stability of colloidal solution is attributed largely to the electric charge of the dispersed particles. This charge causes them to be coagulated or precipitated. On addition of small amount of electrolytes, the ions carrying oppiste charge are adsorbed by sol particles resulting in the neutralization of their charge. When the sol particles either with no charge or reduced charge, come closer due to Brownian movement, they coalesce to form bigger particles resulting in their separation from the dispersion medium. This is what is called coagulating or precipitation of the colloidal solution. The coagulating power of the effective ion, which depend on its charge, is expressed in terms of its coagulating value, defined as its minimum concentration (m mol/L) needed to precipitate a given sol. 100 ml each of two sols of AgI, one obtained by adding `AgNO_(3)` to slight excess of KI and another obtained by adding KI to slight excess of `AgNO_(3)`, are mixed together. Then :A. The two sols will stabilize each otherB. The sol particles will acquire more electric chargeC. The sols will coagulate each other mutuallyD. A true solution will be obtained |
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Answer» Correct Answer - C The sols will neutrakise each other so will coagulate each other. |
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| 546. |
A reddish brown sol `("containing" Fe^(3+))` is obtained by:A. the addition of small amount of `FeCl_(3)` solution to freshly prepared `Fe(OH)_(3)` precipitateB. the addition of `Fe(OH)_(3)` to freshly prepared `FeCl_(3)` solutionC. the addition of `NH_(4)OH` to `FeCl_(3)` solution dropwiseD. the addition of `NaOH to `FeCl_(3)` solution dropwise |
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Answer» Correct Answer - A Reddish brown sol is prepared by adding `FeCl_(3)` in `Fe(OH)_(3)` precipitate. |
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| 547. |
Which of the following represents a multimilecular colloidal particles?A. StarchB. A sol of goldC. ProteinsD. Soaps |
| Answer» Correct Answer - B | |
| 548. |
Which of the following represents macromolecular colloidal particles?A. Solution of goldB. CelluloseC. SoapsD. Synthetic detergents |
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Answer» Correct Answer - B Cellulose is macromolecular colloid. |
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| 549. |
When cells of skeletal vacuoles of a frog were placed in a series of `NaCl` solutions of different concentration solution at `25^(@)C`, it was observed microscopically that they remained unchaged in `0.7%` solution, shrank in a more concentrated and swelled in more dilute solution. Water freezes from the `0.7%` salt solutions at `-406^(@)C`. What is the osmotic pressure of the cell cytoplasm at `25^(@)C`. `(K=_(f)=1.86 kg "mol"^(-1) K)` |
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Answer» `DeltaT_(f)=K_(f)xxixxC_(m)` `=K_(f)xx(1+alpha)xx(W/m)xx1000/(W ("in" g))` `:. 0.406=1.86xx(1+alpha)xx(0.7/58.5)xx(1000/99.3)` `:.1+alpha=(0.4xx58.5xx100)/(1.86xx0.7xx1000)` Assuming dilute solution `100 g H_(2)O =100 mL H_(2)O=0.1 L` solutions `pi=ixx(n_(2)/V)xxRT=(1+alpha)xxW_(2)/(Mw_(2))xx1/V xxRT` `=(0.406xx58.5xx100)/(1000xx1.86xx0.7)xx(0.7/58.5)xx(1/0.1)xx0.082xx298` `=5.34` atm |
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| 550. |
`0.48`of a substance is dissolved in `10.6`g of `C_(6)H_(6)`.The freezing point of benzene is lowered by `1.8^(@)C`what will be the mol.wt.of the substance (`K_(f)`for benzene =`5`)A. `250.2`B. `90.8`C. `125.79`D. `102.5` |
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Answer» Correct Answer - C `DeltaT_(f)=(1000.K_(f)w)/(mW)` `m=(1000K_(f)w)/(cancelOT_(f)W)=(1000xx5xx0.48)/(1.8xx10.6)=125.79` |
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